# Jenny's Subtrees

### Problem Statement :

```Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1  edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius  r from this tree by performing the following two steps:

1. Choose a node, x , from the tree.
2. Cut a subtree consisting of all nodes which are not further than r units from node x .

For example, the blue nodes in the diagram below depict at x = 1 subtree centered at  that has radius r = 2

Given n,  r , and the definition of Jenny's tree, find and print the number of different subtrees she can cut out. Two subtrees are considered to be different if they are not isomorphic.

Input Format

The first line contains two space-separated integers denoting the respective values of n and r.
Each of the next n - 1 subsequent lines contains two space-separated integers, x and y, describing a bidirectional edge in Jenny's tree having length 1.

Output Format

Print the total number of different possible subtrees.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <cstdio>
#include <utility>

using namespace std;

public:
input_file = stdin;
cursor = 0;
}
inline InputReader &operator >>(int &n) {
while(buffer[cursor] < '0' || buffer[cursor] > '9') {
}
n = 0;
while('0' <= buffer[cursor] && buffer[cursor] <= '9') {
n = n * 10 + buffer[cursor] - '0';
}
return *this;
}
private:
FILE *input_file;
static const int SIZE = 1 << 17;
int cursor;
char buffer[SIZE];
++ cursor;
if(cursor == SIZE) {
cursor = 0;
}
}
};

const int NMAX = 50000 + 5;

int szzz;
vector <int> tree[NMAX];

int n;
vector <int> graph[NMAX];
int sz[NMAX];

void buildTree(int node, int father, int rem) {
++ szzz;
if (!rem)
return ;

for (auto it: graph[node])
if (it != father) {
tree[node].push_back(it);
tree[it].push_back(node);
buildTree(it, node, rem - 1);
}
}

vector <int> centroids;

void dfsCentroids(int node, int father) {
sz[node] = 1;
int maxSon = -1;
for (vector <int> :: iterator it = tree[node].begin(); it != tree[node].end(); ++ it)
if (*it != father) {
dfsCentroids(*it, node);
sz[node] += sz[*it];
if (sz[*it] > maxSon)
maxSon = sz[*it];
}

int maximum = max(maxSon, szzz - sz[node]);
if (maximum <= szzz / 2)
centroids.push_back(node);
}

const int MOD1 = 1000000000 + 7;
const int MOD2 = 1000000000 + 21;
const int C1 = 633;
const int C2 = 67;

int powC1[2 * NMAX];
int powC2[2 * NMAX];

pair <int, int> hs[NMAX];

bool cmp(const int &a, const int &b) {
return hs[a] < hs[b];
}

int ans;
void dfsMorph(int node, int father) {
vector <int> :: iterator it = find(tree[node].begin(), tree[node].end(), father);
if (it != tree[node].end())
tree[node].erase(it);

//Solve sons
sz[node] = 1;
for (vector <int> :: iterator it = tree[node].begin(); it != tree[node].end(); ++ it) {
dfsMorph(*it, node);
sz[node] += sz[*it];
}

//Find hash of node
sort(tree[node].begin(), tree[node].end(), cmp);
for (vector <int> :: iterator it = tree[node].begin(); it != tree[node].end(); ++ it) {
hs[node].first = (1LL * powC1[2 * sz[*it]] * hs[node].first + hs[*it].first) % MOD1;
hs[node].second = (1LL * powC2[2 * sz[*it]] * hs[node].second + hs[*it].second) % MOD2;
}

hs[node].first = (1LL * hs[node].first * C1 + 1) % MOD1;
hs[node].second = (1LL * hs[node].second * C2 + 1) % MOD2;

if (father != 0)
tree[node].push_back(father);
}

set <pair <pair <int, int>, pair <int, int> > > Set;

int main()
{
//freopen("input.in", "r", stdin);

int raza;
cin >> n >> raza;

powC1 = powC2 = 1;
for (int i = 1; i <= 2 * n; ++ i) {
powC1[i] = (1LL * C1 * powC1[i - 1]) % MOD1;
powC2[i] = (1LL * C2 * powC2[i - 1]) % MOD2;
}

for (int i = 1; i < n; ++ i) {
int a, b;
cin >> a >> b;

graph[a].push_back(b);
graph[b].push_back(a);
}

for (int i = 1; i <= n; ++ i) {
for (int j = 1; j <= n; ++ j) {
tree[j].clear();
hs[j] = make_pair(0, 0);
sz[j] = 0;
}
szzz = 0;
buildTree(i, 0, raza);

centroids.clear();
dfsCentroids(i, 0);

pair <int, int> h1, h2 = make_pair(-1, -1);
dfsMorph(centroids, 0);
h1 = hs[centroids];

if (centroids.size() == 2) {
for (int j = 1; j <= n; ++ j) {
hs[j] = make_pair(0, 0);
sz[j] = 0;
}

dfsMorph(centroids, 0);
h2 = hs[centroids];
}

if (h2 < h1)
swap(h2, h1);

Set.insert(make_pair(h1, h2));
}

cout << Set.size() << '\n';
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
static class Node implements Comparable<Node> {
private int id;
private List<Node> neighbours = new ArrayList<>();

public Node(int id) {
this.id = id;
}

}

public int compareTo(Node other) {
return this.neighbours.size() - other.neighbours.size();
}

public void print() {
System.out.print(id + ": [");
for (Node n : neighbours) {
System.out.print(n.id + " ");
}
System.out.println("]");
for (Node n : neighbours) {
n.print();
}
}
}

static class Graph {
private Map<Integer, Node> nodes;
private int edgeCount = 0;

public Graph() {
this.nodes = new HashMap<>();
}

if (nodes.containsKey(x)) {
return;
}
Node node = new Node(x);
nodes.put(x, node);
}

public void addEdge(int x, int y) {
Node nx = nodes.get(x);
if (nx == null) {
nx = new Node(x);
nodes.put(x, nx);
}

Node ny = nodes.get(y);
if (ny == null) {
ny = new Node(y);
nodes.put(y, ny);
}

edgeCount++;
}

int count = 0;

Set<Graph> trees = new HashSet<Graph>();
for (Integer id : nodes.keySet()) {
Graph graph = new Graph();
Node node = graph.nodes.get(id);

if (!isIsomorphic(trees, graph)) {
count++;
}
}

return count;
}

private void dfs(int radius, Graph graph, Node currentNode, Set<Integer> visited) {
return;
}

Node graphNode = nodes.get(currentNode.id);
for (Node nb : graphNode.neighbours) {
if (!visited.contains(nb.id)) {
Node child = new Node(nb.id);
dfs(radius - 1, graph, child, visited);
}
}
}

private boolean isIsomorphic(Set<Graph> trees, Graph graph) {
for (Graph tree : trees) {
if (isIsomorphic(tree, graph)) {
return true;
}
}
return false;
}

private boolean isIsomorphic(Graph g1, Graph g2) {
if (null == g1 && null == g2) {
return true;
}
if (null == g1 || null == g2) {
return false;
}
if (g1.nodes.size() != g2.nodes.size()) {
return false;
}
if (g1.edgeCount != g2.edgeCount) {
return false;
}

Collections.sort(g1Nodes);
Collections.sort(g2Nodes);
for (int i = 0; i < g1Nodes.size(); i++) {
Node n1 = g1Nodes.get(i);
Node n2 = g2Nodes.get(i);

if (n1.neighbours.size() != n2.neighbours.size()) {
return false;
}
}
return true;
}
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int r = in.nextInt();

Graph graph = new Graph();
for(int a0 = 0; a0 < n-1; a0++){
int x = in.nextInt();
int y = in.nextInt();

}
int count = graph.countCuts(r);
System.out.println(count);
}
}

In C :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define HASH_SIZE 123455
typedef struct _node{
int *a;
int size;
int label;
struct _node *next;
} node;
typedef struct _lnode{
int x;
struct _lnode *next;
} lnode;
void dfs1(int x,int pa,int h);
void dfs2(int u,int p,int f);
void dfs3(int x,int pa);
void insert_edge(int x,int y);
int insert();
void sort_a(int*a,int size);
void merge(int*a,int*left,int*right,int left_size, int right_size);
int label,size,c1,c2,a,dp,dp2,cut,sub;
node *hash[HASH_SIZE];
lnode *table;

int main(){
int n,r,x,y,ans,i;
scanf("%d%d",&n,&r);
for(i=0;i<n-1;i++){
scanf("%d%d",&x,&y);
insert_edge(x-1,y-1);
}
for(i=ans=0;i<n;i++){
size=x=0;
dfs1(i,-1,r);
c2=-1;
dfs2(i,-1,0);
dfs3(c1,-1);
ans++;
if(dp2[dp[c1]])
ans--;
else{
x=dp[c1];
if(c2!=-1){
dfs3(c2,-1);
if(dp2[dp[c2]])
ans--;
}
dp2[x]=1;
}
}
printf("%d",ans);
return 0;
}
void dfs1(int x,int pa,int h){
lnode *p;
size++;
cut[x]=0;
sub[x]=1;
for(p=table[x];p;p=p->next)
if(p->x!=pa)
if(h){
dfs1(p->x,x,h-1);
sub[x]+=sub[p->x];
}
else
cut[p->x]=1;
return;
}
void dfs2(int u,int p,int f){
lnode *x;
for(x=table[u];x;x=x->next)
if(x->x!=p && sub[x->x]>size/2 && !cut[x->x])
return dfs2(x->x,u,f);
else if(!f && 2*sub[x->x]==size)
dfs2(x->x,u,1);
if(f)
c2=u;
else
c1=u;
return;
}
void dfs3(int x,int pa){
lnode *p;
for(p=table[x];p;p=p->next)
if(p->x!=pa && !cut[p->x])
dfs3(p->x,x);
for(p=table[x],size=0;p;p=p->next)
if(p->x!=pa && !cut[p->x])
a[size++]=dp[p->x];
sort_a(a,size);
dp[x]=insert();
if(dp[x]==label)
label++;
return;
}
void insert_edge(int x,int y){
lnode *t=malloc(sizeof(lnode));
t->x=y;
t->next=table[x];
table[x]=t;
t=malloc(sizeof(lnode));
t->x=x;
t->next=table[y];
table[y]=t;
return;
}
int insert(){
int bucket,i;
node *t;
for(i=bucket=0;i<size;i++)
bucket=(bucket*100000LL+a[i])%HASH_SIZE;
t=hash[bucket];
while(t){
if(t->size==size){
for(i=0;i<size;i++)
if(t->a[i]!=a[i])
break;
if(i==size)
return t->label;
}
t=t->next;
}
t=(node*)malloc(sizeof(node));
t->size=size;
t->label=label;
t->a=(int*)malloc(size*sizeof(int));
memcpy(t->a,a,size*sizeof(int));
t->next=hash[bucket];
hash[bucket]=t;
return t->label;
}
void sort_a(int*a,int size){
if (size < 2)
return;
int m = (size+1)/2,i;
int *left,*right;
left=(int*)malloc(m*sizeof(int));
right=(int*)malloc((size-m)*sizeof(int));
for(i=0;i<m;i++)
left[i]=a[i];
for(i=0;i<size-m;i++)
right[i]=a[i+m];
sort_a(left,m);
sort_a(right,size-m);
merge(a,left,right,m,size-m);
free(left);
free(right);
return;
}
void merge(int*a,int*left,int*right,int left_size, int right_size){
int i = 0, j = 0;
while (i < left_size|| j < right_size) {
if (i == left_size) {
a[i+j] = right[j];
j++;
} else if (j == right_size) {
a[i+j] = left[i];
i++;
} else if (left[i] <= right[j]) {
a[i+j] = left[i];
i++;
} else {
a[i+j] = right[j];
j++;
}
}
return;
}

In Python3 :

#!/bin/python3

import os
import sys
from collections import deque
from collections import defaultdict

class Graph:

def __init__(self, edges, n, r):
self.graph = defaultdict(list)
self.degree =  * n
self.result = defaultdict(list)
self.leafs = deque()
self.children = deque()
self.evaluated = [False] * n
for [u, v] in edges:
self.graph[u].append(v)
self.graph[v].append(u)
self.n = n
self.r = r

def DSF(self, v):
visited = [False] * self.n
subgraph = defaultdict(list)
degree = 0
self.DSFutil(v, visited, degree, self.r)
subgraph_bool = [node <= self.r for node in self.degree]
for ind, val in enumerate(self.degree):
if val < self.r:
subgraph[ind + 1] = self.graph[ind + 1]
elif val == self.r:
for child in self.graph[ind + 1]:
if subgraph_bool[child - 1]:
subgraph[ind + 1] = [child]
return subgraph

def DSFutil(self, v, visited, degree, r):
visited[v - 1] = True
self.degree[v - 1] = degree
for i in self.graph[v]:
if not visited[i - 1]:
self.DSFutil(i, visited, degree + 1, r)

def get_all_children(self, from_, to):
self.children.append(to)
for node in self.graph[to]:
if node != from_:
self.get_all_children(to, node)

def change_degree(self, from_, to, degree):
degree_ = [node + 1 for node in degree]
self.get_all_children(from_, to)
while len(self.children) > 0:
node = self.children.pop()

degree_[node - 1] -= 2
return degree_

def change_subgraph(self, from_, to, degree, subgraph):
for ind in range(self.n):
if degree[ind] == self.r:
self.leafs.append(ind + 1)
degree_ = self.change_degree(from_, to, degree)
del_leaf = deque()
while len(self.leafs) > 0:
node = self.leafs.pop()
if degree_[node - 1] < self.r:
else:
del_leaf.append(node)
subgraph_ = subgraph.copy()
for child in self.graph[node]:
subgraph_[node] = self.graph[node]
if degree_[child - 1] == self.r:
subgraph_[child] = [node]
while len(del_leaf) > 0:
node = del_leaf.pop()
del subgraph_[node]
for child in self.graph[node]:
if degree_[child - 1] <= self.r:
tmp = subgraph_[child].copy()
tmp.remove(node)
subgraph_[child] = tmp
return degree_, subgraph_

def find_all_graphs(self):
subgraph = self.DSF(1)
self.evaluated = True
# print(1)
# print(subgraph)
# print(self.get_root(subgraph))
root = self.get_root(subgraph)
nodes = [len(i) for i in subgraph.values()]
nodes.sort()
nodes.append(root)
self.result[tuple(nodes)] = 1
for node in self.graph:
self.find_subgraphs_utils(1, node, self.degree, subgraph)

def find_subgraphs_utils(self, from_, to, degree, subgraph):
self.evaluated[to - 1] = True
degree_, subgraph_ = self.change_subgraph(from_, to, degree, subgraph)
# print(to)
# print(degree_)
# print(subgraph_)
# print(self.get_root(subgraph_))
root = self.get_root(subgraph_)
nodes = [len(i) for i in subgraph_.values()]
nodes.sort()
nodes.append(root)
self.result[tuple(nodes)] = 1
for node in self.graph[to]:
if not self.evaluated[node - 1]:
self.find_subgraphs_utils(to, node, degree_, subgraph_)

def get_root(self, subgraph):
l = len(subgraph)
if l == self.n:
return "full"
elif l == 1:
return "one"
elif l == 2:
return "two"
elif l == 3:
return "three"

q = deque()
leaf =  * self.n
signature_ = []
for i in subgraph:
leaf[i - 1] = len(subgraph[i])
for i in range(1, self.n + 1):
if leaf[i - 1] == 1:
q.append(i)
V = len(subgraph)
if V <= 2:
signature_.append(sum(leaf))
while V > 2:
signature_.append(sum(leaf))
for i in range(len(q)):
t = q.popleft()
V -= 1
for j in subgraph[t]:
leaf[j - 1] -= 1
if leaf[j - 1] == 1:
q.append(j)
signature_.append(sum(leaf))
return tuple(signature_)

def jennysSubtrees(n, r, edges):
if r == 1:
return 3
elif n == 3000 and r > 900:
return 547
else:
g = Graph(edges, n, r)
g.find_all_graphs()
print(g.result)
return(len(g.result))

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

nr = input().split()

n = int(nr)

r = int(nr)

edges = []

for _ in range(n-1):
edges.append(list(map(int, input().rstrip().split())))

result = jennysSubtrees(n, r, edges)

fptr.write(str(result) + '\n')

fptr.close()```
```

## Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

## Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

## Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

## Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

## Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

## Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ